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A 200 ml buffer solution is made up of 0.20 M HClO and 0.15 M NaClO....

A 200 ml buffer solution is made up of 0.20 M HClO and 0.15 M NaClO. What is the pH after 1.71 grams of barium hydroxide is added to solution? (assume no volume change with addition)? HINTS: Ka of HClO is 2.9x108 and barium hydroxide has a molar mass of 171.35 g/mole.

Expert Solution

First calculate moles of HClO, NaClO & Ba(OH) ₂

We know that, Molarity = No.of moles of solute / volume of solution in L

No. of moles of solute = Molarity volume of solution in L

No. of moles of HClO = 0.20 mol / L 0.200 L = 0.040 mol

No. of moles of NaClO = 0.15 mol / L 0.200 L = 0.030 mol

We have, no. of moles = Mass / Molar mass

no. of moles of Ba(OH) ₂ = 1.71 g / 171.35 g/mol = 0.00998 mol

Ba(OH) ₂ is a strong base. It dissociates completely into ions when dissolved in water.

Ba(OH) ₂ (aq) Ba ²⁺ (aq) + 2 OH ^- (aq)

From reaction, 1 mol Ba(OH) ₂ 2 mol OH ^-

no. of moles of OH ^- = 2 ( no. of moles of Ba(OH) ₂ )

no. of moles of OH ^- = 2 ( 0.00998 mol ) = 0.01996 mol

Consider reaction of hydroxide ions with buffer solution.

HClO (aq) + OH ^- (aq) ClO ^- (aq) + H₂O (l)

Let's use ICE table.

moles	HClO	OH ^-	ClO ^-
I	0.040	0.01996	0.030
C	-0.01996	-0.01996	+0.01996
E	0.02004	0.00000	0.04996

pH of buffer solution is calculated by using Henderson's equation.

pH = pKa + log [ salt ] / [ Acid ]

pH = - log Ka + log [ ClO ^-] / [ HClO ]

pH = - log ( 2.9 10 ^-8 ) + log ( 0.04996 mol / 0.200 L ) / ( 0.02004 mol / 0.200 L)

pH =7.54 + log ( 0.04996 mol / 0.200 L ) / ( 0.02004 mol / 0.200 L)

pH = 7.54 + 0.3967

pH = 7.937

ANSWER : pH of buffer solution after addition of 1.71 g barium hydroxide is 7.94

queen_honey_blossom answered 2 years ago

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